LeetCode（63）Unique Paths II

题目

Follow up for “Unique Paths”:

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[

[0,0,0],

[0,1,0],

[0,0,0]

]

The total number of unique paths is 2.

Note: m and n will be at most 100.

分析

带障碍的路径数计算问题，这个题目与上一题的区别之处在于，m*n的矩阵中有部分设置了障碍，当然有障碍的地方不能通过；

虽然设立了障碍，该题目的本质仍然是一个动态规划问题，我们只需要增加判断当前点是否有障碍的代码即可，若有障碍那么此处不能通行，自然f(i,j)=0，对于其他点，依然用上一题目的推导公式即可！

AC代码

//直接用非递归算法求解
class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {

        if (obstacleGrid.empty())
            return 0;

        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();

        vector<vector<int> > ret(m, vector<int>(n, 0));

        //矩阵首列
        for (int i = 0; i < m; i++)
        {
            //无障碍，则有一条路径，否则不通
            if (obstacleGrid[i][0] != 1)
                ret[i][0] = 1;
            else
                break;
        }//for

        //矩阵首行
        for (int j = 0; j < n; j++)
        {
            //无障碍，则有一条路径，否则不通
            if (obstacleGrid[0][j] != 1)
                ret[0][j] = 1;
            else
                break;
        }//for

        //其余位置
        for (int i = 1; i < m; i++)
        {
            for (int j = 1; j < n; j++)
            {
                //当前位置为障碍，则到此处路径数为0
                if (obstacleGrid[i][j] == 1)
                    ret[i][j] = 0;
                else{
                    ret[i][j] = ret[i][j - 1] + ret[i - 1][j];
                }//else
            }//for
        }//for

        return ret[m - 1][n - 1];
    }//uniques

};

GitHub测试程序源码